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René Gy Rene. Gy@ Numericable #A67 INTEGERS 20 (2020) BERNOULLI-STIRLING NUMBERS Ren´eGy [email protected] Received: 5/29/19, Revised: 2/29/20, Accepted: 8/13/20, Published: 8/31/20 Abstract Congruences modulo prime powers involving generalized Harmonic numbers are known. While looking for similar congruences, we have encountered a curious tri- angular array of numbers indexed with positive integers n; k, involving the Bernoulli and cycle Stirling numbers. These numbers are all integers and they vanish when n − k is odd. This triangle has many similarities with the Stirling triangle. In particular, we show how it can be extended to negative indices and how this ex- tension produces a second kind of such integers which may be considered as a new generalization of the Genocchi numbers and for which a generating function is easily obtained. But our knowledge of these integers remains limited, especially for those of the first kind. 1. Introduction Let n and k be non-negative integers and let the generalized Harmonic numbers (k) (k) Hn , Gn be defined as n X 1 H(k) := ; n jk j=1 (k) X 1 Gn := , i1i2 ··· ik 1≤i1<i2<··<ik≤n (1) (1) Pn 1 (0) (0) with Hn = Gn = j=1 j = Hn = Gn; Hn = n and Gn = 1. We have [10] n + 1 = n!G(k); k + 1 n n k being the cycle Stirling number (or unsigned Stirling number of the first kind), so that the Harmonic and cycle Stirling numbers are inter-related by the convolution k−1 n + 1 X n + 1 k = − (−1)k−jH(k−j) , (1.1) k + 1 n j + 1 j=0 2 which is obtained as a direct application of the well-known relation between ele- mentary symmetric polynomials and power sums [9]. Extended congruences for the (k) Harmonic numbers Hp−1, modulo any power of a prime p are known [16], [7]. Our initial motivation for the work reported in the present paper is to look for similar (k) congruences modulo prime powers, involving Gp−1, or the cycle Stirling numbers p (k) (k) k+1 , instead of Hp−1. We will show that such similar congruences for Gp−1 do exist, but that they are just the particular prime instances of not very well-known but elementary identities for the cycle Stirling numbers. This will lead us to in- troduce a triangular array of integers, the Bernoulli-Stirling numbers, involving the Bernoulli and Stirling numbers which we believe is new. 2. Notation and Preliminaries In addition to what was exposed in the previous introduction, further notation that we use throughout this paper is presented in this section, along with classical results which we will need. Most of these results can be found in textbooks like [6] and they are given hereafter without proof. In the following, p denotes a prime number, x denotes the argument of a gener- ating function, of a polynomial function or of a formal power series, [[xn]]f(x) the xn m coefficient of n! in f(x) and D f(x) is the m-order derivative of f(x) with respect to x. We will use the Iverson bracket notation: P = 1 when proposition P is true, and P = 0 otherwise. For q a rational number, we denote by vp(q) the p-adic order of q and we have vp(q1 · q2) = vp(q1) + vp(q2) and vp(q1 + q2) ≥ min (vp(q1); vp(q2)). n P n k n The binomial coefficients k are defined by k k x = (1 + x) , whatever the sign of integer n. They obviously vanish when k < 0. They are easily obtained n n−1 n−1 by the basic recurrence relation k = k + k−1 . When n > 0, we have −n kn+k−1 k = (−1) n−1 and we have the well-known inversion formula X kn (−1)k−j = [n = j]: (2.1) j k k≥0 n The cycle Stirling numbers k , n ≥ 0 may be defined by the horizontal generating function n−1 X n Y xk = (x + j); (2.2) k k≥0 j=0 where an empty product is meant to be 1. Alternatively, they have the exponential generating function k k X nxn (−1) ln(1 − x) = : (2.3) k n! k! n≥0 3 They obviously vanish when k < 0 and k > n. They are easily obtained by the n n−1 n−1 0 basic recurrence k = (n − 1) k + k−1 , valid for n ≥ 1, with k = [k = 0]. They also obey the generalized recurrence relation n + 1 X h + m n = : (2.4) m + 1 m h + m h≥0 n We let k , n ≥ 0, be the partition Stirling numbers (or Stirling numbers of the second kind). They also vanish when k < 0 and k > n. Their basic recurrence is n n−1 n−1 0 k = k k + k−1 for n ≥ 1, with k = [k = 0]. They have the following exponential generating function X nxn (ex − 1)k = : (2.5) k n! k! n≥0 We will make use of the following Lemma about the Stirling numbers, for which there is a proof in [6], p. 266-271. Lemma 2.1. Let n; k be non-negative integers. There exists a polynomial Qk 2 n Q[X] of degree 2k, such that Qk(n) coincides with n−k . If k > 0 then 0; 1; ···; k are n+k roots of the polynomial Qk. Moreover, we have Qk(−n) = n and in particular Qk(−1) = 1. For n ≥ 0, let Bn be the Bernoulli numbers. The first of them are B0 = 1;B1 = 1 1 1 − 2 ;B2 = 6 ;B3 = 0;B4 = − 30 ; ::: and for h > 0, we have B2h+1 = 0. They have the well-known exponential generating function t X tn = B (2.6) et − 1 n n! n≥0 and they obey the recurrence n X n (−1)nB = B : (2.7) n k k k=0 We will also make use of the Von Staudt-Clausen theorem which states that the denominator of Bn, in reduced form, is the product of all primes p such that p − 1 divides n. In particular, any prime may divide the denominator of a Bernoulli number once at most. 3. Two Identities for the Cycle Stirling Numbers In this section, we re-demonstrate two identities for the cycle Stirling numbers which, in spite of their similarity to Equation (2.4), do not seem to be very well- known. 4 Theorem 3.1. Let m; n be non-negative integers. We have n + 1 X h + m n = (−1)n−m (−n)h. (3.1) m + 1 m h + m h≥0 Moreover, if n > 0, n X h + m − 1 n = (−1)n−m (−n)h. (3.2) m m − 1 h + m h≥0 In more compact and symmetric formulations, these two identies also read n + 1 X h n (−1)n−m (−n)m = (−n)h (3.3) m + 1 m h h and, if n > 0, n X h − 1 n (−1)n−m (−n)m = (−n)h. (3.4) m m − 1 h h Remark. These identities are not in [6] where quite many finite sums, recurrences and convolutions involving Stirling numbers are reported. Our Equation (3.1) may be obtained as a particular case of Theorem 3 in [2]. An identity equivalent to our Equation (3.2), is obtained incidentally in [1], where it is not even labelled. Another identity, equivalent to our Equation (3.2), is the equation (18) from [15], where it is said to be new. Proof of Theorem 3.1. Like in [1], our proof will highlight that Equation (3.1) and Equation (3.2) are actually closely related to the convolution Equation (1.1) Qn−1 between the Harmonic and cycle Stirling numbers. Let fn(x) := h=0(x − h). We are going to show that, for m ≥ 1, m m−1 h n−1 D fn(x) X D fn(x) X 1 m = − (−1)m−h . (3.5) m! h! (x − j)m−h h=0 j=0 Pn−1 Q Pn−1 1 It is true for m = 1, since Dfn(x) = j=0 h6=j(x − h) = fn(x) j=0 x−j . We suppose (induction hypothesis) that it is true for some m, so that Dm+1f (x) Dmf (x) 1 Dmf (x) (m + 1) n = D n = D m n (m + 1)! m! m m! 5 0 1 m+1 m−1 h+1 n−1 D fn(x) 1 X D fn(x) X 1 (m + 1) = − (−1)m−h (m + 1)! m @ h! (x − j)m−h A h=0 j=0 0 1 m−1 h n−1 1 X D fn(x) X 1 + (−1)m−h (m − h) m @ h! (x − j)m+1−h A h=0 j=0 0 1 m h n−1 1 X D fn(x) X 1 = (−1)m−h m @ (h − 1)! (x − j)m+1−h A h=1 j=0 0 1 m−1 h n−1 1 X D fn(x) X 1 + m (−1)m−h m @ h! (x − j)m+1−h A h=0 j=0 0 1 m−1 h n−1 1 X D fn(x) X 1 − (−1)m−h m @ (h − 1)! (x − j)m+1−h A h=1 j=0 0 1 m n−1 1 D fn(x) X 1 = m @ (m − 1)! x − j A j=0 0 1 m−1 h n−1 1 X D fn(x) X 1 + m (−1)m−h m @ h! (x − j)m+1−h A h=0 j=0 m h n−1 X D fn(x) X 1 = (−1)m−h .
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